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Angular Momentum Angular Momentum Daniel Wysocki and Nicholas Jira April 2, 2015 Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 1 / 30 Introduction Introduction Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 2 / 30 Introduction Quantum Numbers the stationary states of the hydrogen atom are given by three numbers, n, `, and m n is the principal quantum number, and determines the energy of the state ` and m are related to the orbital angular momentum Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 3 / 30 Introduction Angular Momentum classically, a particle’s angular momentum is given by ypz − zpy L = r × p = zpx − xpz xpy − ypx now we simply replace classical momentum with the quantum momentum operator y ∂/∂z − z ∂/∂y ı ı L = z ∂/∂x − x ∂/∂z = (r × ∇) ~ x ∂/∂y − y ∂/∂x ~ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 4 / 30 Eigenvalues Eigenvalues Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 5 / 30 Eigenvalues Fundamental Commutation Relations Lx and Ly do not commute [Lx , Ly] = [ypz − zpy, zpx − xpz ] = [ypz , zpx ] − [ypz , xpz ] − [zpy, zpx ] + [zpy, xpz ] the only terms which fail to commute are [x, px ], [y, py], and [z, pz ] [Lx , Ly] = ypx [pz , z] + xpy[z, pz ] = ı~(xpy − ypx ) = ı~Lz [Lx , Ly] = ı~Lz ;[Ly, Lz ] = ı~Lx ;[Lz , Lx ] = ı~Ly Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 6 / 30 Eigenvalues Uncertainty Principle 1 2 σ2 σ2 ≥ h[A, B]i A B 2ı 1 2 2 σ2 σ2 ≥ hı L i = ~ hL i2 Lx Ly 2ı ~ z 4 z σ σ ≥ ~|hL i| Lx Ly 2 z Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 7 / 30 Eigenvalues Total Angular Momentum since Lx and Ly do not commute, there are no eigenfunctions of both Lx and Ly however, the square of the total angular momentum does commute with Lx 2 2 2 2 L = L · L = Lx + Ly + Lz 2 2 2 [L , Lx ] = 0;[L , Ly] = 0;[L , Lz ] = 0 or [L2, L] = 0 Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 8 / 30 Eigenvalues Ladder Operator since L2 is compatible with each component of L, we can hope to find simultaneous eigenstates of L2 and any given component, say Lz 2 L f = λf and Lz f = µf we define the ladder operator L± ≡ Lx ± ıLy [Lz , L±] = [Lz , Lx ] ± ı[Lz , Ly] = ı~Ly ± ı(−ı~Lx ) = ±~(Lx ± ıLy) 2 [Lz , L±] = ±~L± and [L , L±] = 0 Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 9 / 30 Eigenvalues Ladder Operator and Eigenfunctions 2 if f is an eigenfunction of L and Lz , so too is L±f 2 since L and L± commute, 2 2 L (L±f ) = L±(L f ) = L±(λf ) = λ(L±f ) 2 L±f is an eigenfunction of L with eigenvalue λ since [Lz , L±] = ±~L±, Lz (L±f ) = (Lz L± − L±Lz )f + L±Lz f = ±~L±f + L±(µf ) = (µ ± ~)(L±f ) so L±f is an eigenfunction of Lz with eigenvalue µ ± ~ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 10 / 30 Eigenvalues Raising and Lowering Operators L±f is an eigenfunction of Lz with eigenvalue µ ± ~ L+ is the “raising” operator, since it increases the eigenvalue of Lz by ~ L− is the “lowering” operator, since it decreases the eigenvalue of Lz by ~ for a given λ, we obtain a “ladder” of states, with each “rung” separated from its neighbors by ~ in the eigenvalue of Lz Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 11 / 30 Eigenvalues Top Rung 2 2 2 2 L = Lx + Ly + Lz if we allowed the raising operator to be applied forever, eventually 2 we would reach a point where Lz > L , which cannot be there must exist a “top rung” of the ladder, ft, such that L+ft = 0 let ~` be the eigenvalue of Lz at this top rung 2 Lz ft = ~`ft; L ft = λft Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 12 / 30 Eigenvalues Top Rung now we investigate what happens when one ladder operator is applied to its inverse 2 2 L±L∓ = (Lx ± ıLy)(Lx ∓ ıLy) = Lx + Ly ∓ ı(Lx Ly − LyLx ) 2 2 = L − Lz ∓ ı(ı~Lz ) solving for L2 gives 2 2 L = L±L∓ + Lz ∓ ~Lz Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 13 / 30 Eigenvalues Top Rung we use the bottom of the ±, and find that 2 2 2 2 2 2 L ft = (L−L+ + Lz + ~Lz )ft = (0 + ~ ` + ~ `)ft = ~ `(` + 1)ft 2 2 2 L ft = ~ `(` + 1)ft = λft =⇒ λ = ~ `(` + 1) so we have found the eigenvalue of L2 in terms of the maximum eigenvalue of Lz Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 14 / 30 Eigenvalues Bottom Rung 2 2 2 2 L = Lx + Ly + Lz for the same reasons, there must exist a bottom rung, fb, such that L−fb = 0 ¯ let ~` be the eigenvalue of Lz at this bottom rung ¯ 2 Lz fb = ~`fb; L fb = λfb Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 15 / 30 Eigenvalues Bottom Rung we now use the top of the ±, where we had previously used the bottom, and find that 2 2 2 ¯2 2 ¯ 2 ¯ ¯ L fb = (L+L− + Lz − ~Lz )fb = (0 + ~ ` − ~ `)fb = ~ `(` − 1)fb 2 2 ¯ ¯ 2 ¯ ¯ L fb = ~ `(` − 1)fb = λfb =⇒ λ = ~ `(` − 1) Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 16 / 30 Eigenvalues Combining the Top and Bottom we see that 2 2 λ = ~ `(` + 1) = ~ `¯(`¯− 1) =⇒ `(` + 1) = `¯(`¯− 1) there are two possibilities here 1 `¯= ` + 1 that would mean the bottom rung is higher than the top! 2 `¯= −` Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 17 / 30 Eigenvalues Eigenvalues of Angular Momentum we have just shown that the eigenvalues of Lz are m~, where m = −`, −` + 1,..., 1 + `, +` if we let the number of eigenvalues be N , then ` = −` + N ` = N /2 ` must be an integer, or a half-integer ` = 0, 1/2, 1, 3/2,... the eigenfunctions are characterized by ` and m 2 m 2 m m m L f` = ~ `(` + 1)f` ; Lz f` = ~mf` Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 18 / 30 Eigenfunctions Eigenfunctions Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 19 / 30 Eigenfunctions Angular Momentum in Spherical Coordinates the angular momentum operator is ı L = (r × ∇) ~ in spherical coordinates, the gradient is given by ∂ 1 ∂ 1 ∂ ∇ =r ˆ + θˆ + φˆ ∂r r ∂θ r sin θ ∂φ r is simply rrˆ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 20 / 30 Eigenfunctions Angular Momentum in Spherical Coordinates ∂ ∂ 1 ∂ L = ~ r(ˆr × rˆ) + (ˆr × θˆ) + (ˆr × φˆ) ı ∂r ∂θ sin θ ∂φ (ˆr × rˆ) = 0, (ˆr × θˆ) = φˆ, and (ˆr × φˆ) = −θˆ ∂ 1 ∂ L = ~ φˆ − θˆ ı ∂θ sin θ ∂φ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 21 / 30 Eigenfunctions Angular Momentum in Spherical Coordinates write the unit vectors θˆ and φˆ in cartesian coordinates θˆ = (cos θ cos φ)ˆı + (cos θ sin φ)ˆ − (sin θ)kˆ φˆ = −(sin φ)ˆı + (cos φ)ˆ " ∂ L = ~ (− sin φˆı + cos φˆ) ı ∂θ # 1 ∂ − (cos θ cos φˆı + cos θ sin φˆ− sin θkˆ) sin θ ∂φ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 22 / 30 Eigenfunctions Angular Momentum in Spherical Coordinates separating the x, y, and z components, we find ∂ ∂ L = ~ − sin φ − cos φ cot θ x ı ∂θ ∂φ ∂ ∂ L = ~ + cos φ − sin φ cot θ y ı ∂θ ∂φ ∂ L = ~ z ı ∂φ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 23 / 30 Eigenfunctions Ladder Operators in Spherical Coordinates now we consider the ladder operators ∂ ∂ L = L ±ıL = ~ (− sin φ ± ı cos φ) − (cos φ ± ı sin φ) cot θ ± x y ı ∂θ ∂φ by Euler’s formula, cos φ ± ı sin φ = exp(±ıφ) ∂ ∂ L = ± exp(±ıφ) ± ı cot θ ± ~ ∂θ ∂φ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 24 / 30 Eigenfunctions Ladder Operators in Spherical Coordinates ! ∂2 ∂ ∂2 ∂ L L = − 2 + cot θ + cot2 θ + ı + − ~ ∂θ2 ∂θ ∂φ2 ∂φ 2 2 recall L = L±L∓ + Lz ∓ ~Lz " # 1 ∂ ∂ 1 ∂2 L2 = − 2 sin θ + ~ sin θ ∂θ ∂θ sin2 θ ∂φ2 Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 25 / 30 Eigenfunctions Eigenfunctions of L2 2 m now we apply L to its eigenfunction, f` (θ, φ), which has 2 eigenvalue ~ `(` + 1) " # 1 ∂ ∂ 1 ∂2 L2f m = − 2 sin θ + f m = 2`(` + 1)f m ` ~ sin θ ∂θ ∂θ sin2 θ ∂φ2 ` ~ ` this is simply the angular equation ∂ ∂y ∂2Y sin θ sin θ + = −`(` + 1) sin2 θY ∂θ ∂θ ∂φ2 Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 26 / 30 Eigenfunctions Eigenfunctions of Lz m f` is also an eigenfunction of Lz with eigenvalue m~ ∂ L f m = ~ f m = mf m z ` ı ∂φ ` ~ ` this is equivalent to the azimuthal equation 1 d2Φ = −m2 Φ dφ2 Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 27 / 30 Eigenfunctions Spherical Harmonics m m f` is simply Y` (θ, φ), the spherical harmonic (after normalization) 2 spherical harmonics are eigenfunctions of L and Lz when solving the Schrödinger equation by separation of variables, we “inadvertently” constructed eigenfunctions of the three commuting operators 2 2 H ψ = Eψ; L ψ = ~ `(` + 1)ψ; Lz ψ = ~mψ Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 28 / 30 Eigenfunctions Schrödinger Equation " !# 2 1 ∂ ∂ψ 1 ∂ ∂ψ 1 ∂2ψ − ~ r 2 + sin θ + 2m r 2 ∂r ∂r r 2 sin θ ∂θ ∂θ r 2 sin2 θ ∂θ2 +V ψ = Eψ we can now write the Schrödinger equation in this form 1 ∂ ∂ − 2 r 2 + L2 ψ + V ψ = Eψ 2mr 2 ~ ∂r ∂r Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 29 / 30 Eigenfunctions Thank You Daniel Wysocki and Nicholas Jira Angular Momentum April 2, 2015 30 / 30.
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